Question 771842
<pre>
To make it easy, let's change feet to "dekafeet".

1 dekafoot = 10 feet, the length of one of those measuring chains
used at a football game to see if a first down was made.  So 1
of those chains is 1 dekafoot long. 

So the problem is:

how high is a semi elliptical arch of height 2 dekafeet. and span 
8 dekafeet at a distance of 1.5 dekafeet from the end?  So we'll 
draw the arch:

We'll let 4 of the 8 dekafeet be on the left of the origin, and 4 on
the right, making the vertices (-4,0) and (4,0)

{{{drawing(400,240,-5,5,-3,3,
graph(400,240,-5,5,-3,3), red(arc(0,0,8,-4,0,180)) )}}}

The whole ellipse looks like this:

{{{drawing(400,240,-5,5,-3,3,
graph(400,240,-5,5,-3,3,

(-sqrt(16-x^2)/2)*(sqrt(sin(10x))/sqrt(sin(10x)))), 
red(arc(0,0,8,-4,0,180)),locate(1.5,.4,a=4), locate(.2,1,b=2) 


)}}} 

Since a=4 and b=2, the equation

{{{x^2/a^2}}}{{{""+""}}}{{{y^2/b^2}}} {{{""=""}}} 1

{{{x^2/4^2}}}{{{""+""}}}{{{y^2/2^2}}} {{{""=""}}} 1

{{{x^2/16}}}{{{""+""}}}{{{y^2/4}}} {{{""=""}}} 1

Let's solve for y, so we can find the height 1.5 dekafeet
from the the end.

Clear of fractions by multiplying every term by 16

x² + 4y² = 16

     4y² = 16 - x²
      
      y² = {{{(16-x^2)/4}}}

       y = {{{"" +- sqrt(16-x^2)/2}}}

The + is the arch, the - is the dotted line lower part, below
the ground.

1.5 dekafeet from the right end is 4-1.5 = 2.5 dekafeet from the
origin, see the green line

{{{drawing(400,240,-5,5,-3,3,green(line(2.5,1.56,2.5,0)),
graph(400,240,-5,5,-3,3,

(-sqrt(16-x^2)/2)*(sqrt(sin(10x))/sqrt(sin(10x)))), 
red(arc(0,0,8,-4,0,180)),locate(1.5,.4,a=4), locate(.2,1,b=2) 


)}}}  

So we substitute x=2.5in

       y = {{{ sqrt(16-x^2)/2 }}}

       y = {{{ sqrt( 16-(2.5)^2 )/2}}}
      
       y = {{{sqrt(9.75)/2}}} = 1.5612495 dekafeet.

So the answer in feet is 15.612495.

Edwin</pre>